Suppose the step function is defined as follows. A function $f$ is a step function on $[a,b]$, if there exists a finite partition $P$ of $[a,b]$ such that $f$ is constant on the interior of each subinterval $I_k$ of $P$, where $I_k = [x_{k-1}, x_k]$ for $k=1,\dotsc,n$. Also suppose $a = x_0$ and $b = x_n$.
Questions:
(a) Since the function is constant on the interior of each subinterval, discontinuities occur at the endpoints. Even though there are finitely many endpoints, is it right to say that the value of $f$ at these endpoints may not be finite?
(b) If the value of $f$ at these endpoints are finite, then it is possible to show that it is Riemann integrable. But if they are not finite, would the step function also be Riemann integrable?
(c) A related question: Based on the above definition of the step function, is $f$ bounded on $[a,b]$?
You don't say this, so to avoid foolishness: suppose throughout $a < b$.
(a) Your definition makes no requirement of $f$ on $\{x_0, x_1, x_2, \dots, x_N\}$, so $f$ is not required to exist or be defined at those points. Since the value of a function (from any set to the reals) is always a number, if it is defined, the value is finite. However, $f$ is free to be undefined at the various $x_i$.
(b) No. The partition of the Riemann sum is not required to be a refinement of the partition $P$ given in the definition. If $f$ is undefined somewhere, say at $x_i$ for some $i \in [0,N]$, then every Riemann sum partition that does not have $x_i$ as a boundary has it interior to some $I_k$ (using the notation in your definition). Consequently, a Riemann sum may sample the undefined point, making the sum undefined. This continues through the forest of partitions all the way through the limit as the diameter of the partition goes to zero. Consequently, no matter how far along the limit of diameter we go, there are still Riemann sums that are undefined -- this obstructs the existence of the limit. Consequently, the limit does not exist and such a step function is not integrable.
(c) No. $f$ isn't even required to be defined on $[a,b]$. We can ask: Is the restriction of $f$ to the intersection of its domain and $[a,b]$ bounded? Yes. You have $N$ constant segments, say of height $c_1$, ..., $c_N$, and up to $N$ finite values $f(x_1)$, ..., $f(x_N)$. Of course a collection of between $N$ and $2N$ numbers is bounded.